Question

find the derivative of y=(sin(x)sec(x))/csc^2(x) sqrt(csc^2(x)-1

Answer 1

this is what the equationlooks like more clearly\[y=\frac{ \sin(x)\sec(x) }{ \csc^2(x) }\sqrt{\csc^2(x)-1}\]

Answer 2

that's nothing
csc^2(x)=1+tan^2(x)
so csc^2(x)-1=tan^2(x)
and we have also csc^2(x)=1/sin^2(x)

Answer 3

you just need to clean your function little bit
and then differentiate

Answer 4

i mean cot^2(x) not tan sorry lol

Answer 5

1. make sinxsecx/csc^2x to tanxsin^2x

2. do first(derivative2nd)+2nd(derivative1st)
so derivative of tanxsin^2x = tan(2sinxcosx)+sin^2x(sec^2x)
and the square root csc^2x -1 is 1/2(csc^2x-1)^-1/2 *(2cscx)(-cscxcotx)
so its tanxsin^2x(that long retrice2nd one) +the square root one(tan(2sinxcox whatever it is)

Answer 6

this is jut playing with some trig identities that's all

Answer 7

ok but in the fraction would you get (-cosx(secxtanx))/cotx+1?

Answer 8

you have \(\large \rm tan(x)sin^2(x).cot(x)=sin^2(x)\)

Answer 9

if we have \(\large \rm cot^2(x)=csc^2(x)-1 ~\longleftarrow~Pyathagorean~Identity\)
then \(\large \rm cot(x)=\sqrt{csc^2(x)-1}\)
we also know that \(\large \rm sin(x).sec(x)=sin(x).\frac{1}{\Large cos(x)}=\frac{sin(x)}{cos(x)}=tan(x)\)

Answer 10

and also \(\large \rm 1/sec^2(x)=sin^2(x)\) combining every thing together \(\large \rm tan(x).sin^2(x).cot(x)=sin^2(x)\)

Answer 11

Thank you! However I have one more question: were did you get 1/sec^2 and what did you do with the csc^2(X) on the bottom?

Answer 12

oh that's an error on my side i meant 1/csc
not sec

Answer 13

you are welcome